A Level Maths Made Easy - Coding Stats & Mechanics

So today we're going to be looking at standard deviation and coding uh both of which are fairly short topics standard deviation I think is going to be quite quick for you coding might take a bit longer but overall it's not that bad and again as always the worksheet and the mark scheme are both going to be uploaded right after this video so if you would like to have a go at the uh work beforehand you can do that or what I'd recommend doing is having the worksheet out in front of you and then as I do one example attempt the next example yourself then continue the video I'd also recommend a couple of days after watching the video maybe one or two days retrying the questions yourself and seeing how well you do that way you can actually identify if you know enough I would also like to point out that I did get a question before about does this cover everything this covers all the types of question you can be given however you do need to do some practice by yourself the more practices you do the better and these questions are great for that but of course you may want to look at other questions eventually but anyways without further Ado let's just get into it so when we look at standard deviation and coding what is it well let's start off with the coding aspect because that's what's going to come up first in our questions so coding may sound like something that we do in computer science and it is what we do in computer science but coding is just a method by which we simplify data so let's say for example I'm looking into house prices okay so we would say house prices if we look at at the value of house prices you're probably going to get something in the hundreds of thousands region so you might have something like 300,000 then you might have 350,000 you might have 400,000 and you might have 500,000 right those are some stereotypical house prices now imagine if you're actually calculating the median of those values well that means you have to add up all of these numbers and these numbers are quite big right 300,000 plus 350,000 Plus 400,000 + 500,000 what coding does is it allows us to simplify numbers so for example I might divide all of these numbers by 100,000 and now these new numbers are going to just be 3 3.5 4 and five now why do we do this well there's two reasons one reason is that bigger numbers are actually more taxing computationally I'm not just talking about with your brain but when you use calculators or computers when you have very large numbers it actually makes the program run slower and cause it to crash now of course with four numbers that's not going to be a case but for example some of the statistics we have in the UK are things like median house prices that's the price of every house in the country of which there are roughly tens of millions that's where simplifying the numbers will make the numbers go a lot faster you'll be able to calculate the average a lot quicker and then as we'll see later you can actually uncode the median itself and then you'll get what the actual value is so for example if you worked at the median of this number um I probably should put another value to get something roughly about four probably 3.75 actually now what that translates to in real life would be 375,000 which we can do very quickly so instead of using all these massive numbers we can then work out the value using simple numbers and then work out what the actual value is and again that saves time computationally which is really really important the second reason is probably a bit more nuanced with Statistics what do we tend to do with them we tend to show them to other people and those people might not necessarily be mathematicians so showing really large numbers for for example if you're talking about stars the masses of stars or the masses of planets it's very difficult to have these masses which by the way are nor normally look something like this 10.25 * 10 to the^ 24 that number means nothing to the common person it doesn't mean anything in fact it barely means anything to scientists so if we code the data we can make the numbers more manageable so for example if I'm comparing the masses of two planets and I have something like like this if I divide them both by time 10^ of 24 instead I have something like 7.25 and 6.24 which is a lot easier to actually visualize the number to see how much bigger it is okay so those are a few situations where you might do coding and how do you do it well there is it's essentially the same as what I've just done you can divide you can add you can subtract and you can times you can't really square or do anything like that normally we just add subtract times or divide okay and that's really important and as you can see we'll have some kind of formula like this and I'm going to show you what that exactly means so if you're coding data you just sub the values in and work out what the value is nice and simple but you're going to be uncoding as well so I hope that makes sense now what about standard deviation what is the standard deviation and I'm going to show you the um different formulas and everything shortly what is the standard deviation the standard deviation tells you how far on average values are from the center so that might sound really strange but let me visualize this for you let's say we have our values along the bottom and we have the number at that specific value so for example it could be the height of people in ctim you expected to have a bell-shaped curve around the medium which we normally give as a me the standard deviation will tell you effectively how wide this peak is how spread out your data is so it's a new measure of spread which is very very accurate it's more accurate than the range and the inter quarar range okay so just as a visualization if you have something that has a very low standard deviation it's very sharp so low standard deviation low SD you have a very sharp Peak because that width is very small whereas something with a very large standard deviation we'll look something like this can you see how that data is far more spread out so again it's a measure of spread and the higher the standard deviation the greater the spread it's basically saying that on average these values are further apart if they have a high standard deviation low standard deviation means that they're very close together and a standard deviation of zero means that all the values are the mean okay so how exactly do we work it out well the standard deviation is a very interesting formula that you get given but here is the actual formula for it so standard deviation is given by the letter Sigma and it equals the square root of the sum of fx^ 2/ n^2 minus the sum of FX over n s I know it looks really strange right but let's break this down for a second this thing here is literally the same as the mean and this FX and FX s you should recognize from the mean okay so it's the frequency times the um midpoint of the two of the classes so that's actually okay n is the number of pieces of data that you have so overall that's not too bad and I will like I would like to point out that they normally give you these values here the sum of FX squ and the sum of FX and I may even show you later on how to do this all on a calculator Anyway by entering in the data yourself but it's very situational because calculators don't just do it for every single table that exists hopefully that gives you a bit of a brief introduction into what standard deviation and coding are but let's actually get into doing some of these questions okay so let's get on with the first question now you may notice that this is the exact same question as we did in the linear interpolation video however we only did part A and we're going to do part A from scratch as if we haven't done it before just to give us a little bit of revision and again as we go forward this is going to come up a few more times where we've only done part A maybe A and B of A question so in a later video when we learn about the other topics in stats or mechanics we then go through AB again plus also CD and E so this is something you should expect and it's actually a good thing it means we're going to automatically revise previous topics which and I've said this many many times but a level math is very interconnected so please be aware of that so again we have this coded times frequency we need to use linear interpolation to estimate the mean of the median times so all we have to do for that is we start part A we do n/ two so the number of people in this is 50 how did I know that well there's two ways you can either add all of these up or you can just look at the sample size over there and we'll get our calculator ready as well just so we have that available and what it was just done was a for another video uh and we need to actually use linear interpetation so if we do this we're looking for the 25th value and all we do is we go up and so that'll be the first value then there's the first 10 values then there's first 29 values so I know it's in this class over over here and if you remember what you have to do again if you're unsure about what I'm about to do please go back to the previous video in linear interpolation just to get some practice with that so the first thing you do is you take the difference in the actual class that you're looking at so in this case we're going to do 15 - 10 and that is going to be over the median minus 10 because remember the median is somewhere in between 10 and 15 we need to work out what it is then we take the difference in the cumulative frequency so 29 and 10 and the bottom part is the 25th value so 25 - 10 and then you just need to do some rearranging in order to work this out again use just a calculator nice and easy so 29 - 10 over 25 - 10 gives me 19 over 15 on the right hand side so 19 over 15 probably could have done that in my head but oh well 15 - 10 is 5 over Q2 - 10 you can do this a lot faster with the rearranging if you'd like which completely up to if you want to I'm going to times both sides by Q2 - 10 to give it a 195 Q2 - 10 then we're going to Times by 15 and divide by 19 so that would be 15 * 5 over 19 = Q2 - 10 which means Q2 is just if we do it up here Q2 is is equal to 15 * 5 over 19 that is a 19 over here by the way doesn't look like it plus 10 so let's work out what that would be so we have 15 * 5 over 19 plus on another 10 which gives us an answer of 13.94% and if you notice that is in between 10 and 15 so that does make quite a bit of sense and now we get on to the actual new stuff which is standard deviation and the coding so first of all let's start with the standard deviation so that is part B now if you look back at the standard deviation formula we have this so if we just copy this or move this over let's keep it next to us we can look at the kind of stuff that we need so the first thing we need is we need f * X and also f * x^2 and remember this Sigma symbol so it's a capital Sigma means sum of so all you're going to do is add two new columns maybe three if you'd like and the first one is going to just be FX remember frequency is f and X is either the value in this column or it's the midpoint depending on what kind of data you have so so here we have grouped data so we need to find the midpoint however sometimes it's just a single value in which case you don't need to use the midpoint so that's why I said maybe three columns so I personally just like to write it next to it so the midpoint of each of these so 2.5 7.5 remember the method I told you on how to work at the midpoint of any two numbers you just add the numbers together so 0 + 5 and then you divide it by two that automatically gives you the midpoint in case you get numbers that are a bit weird and here will be 12.5 here would be 20 and here would be so this is maybe a slightly trickier one so you just do 25 plus 40 gives you 65 / 2 32.5 okay so either way you want to do it and these values here are our X's so f of x is literally just going to be 1 in this case time 2.5 so it's going to be 2.5 here's going to be 9 * 7.5 in fact to make this slightly easier we're going to actually get rid of the um the numbers from the previous part because we don't need the cumulative frequency anymore so let's just get rid of that or we can uh leave it for a bit later so the next one would be 9 * 7.5 9 * 7.5 67.5 then you have 19 * 12.5 those of you that are perhaps have already spotted a slightly faster way to do this or if you're wondering if there's a faster way to do this I want you to know that you there is and I'm going to show you that a bit later on so this is kind of the longer method of doing it but at the same time it will ensure that you have the cor a correct method because what you don't want to do is try and use a shortcut that you don't fully understand and then end up getting the wrong answer because then it because it defeats the purpose at the end of the day you want the right answer rather than a quick answer okay and what I mean by that is if you remember how calculators work you can actually do all of your calculations in one go so at the end what I like to do is I like to add them all and put them in their own little box at the bottom so all I'm going to do now is add up all of these numbers so we have 2.5 plus 67.5 + 2 37.5 + 280 Plus 2 27.5 it should also be noted that a lot of the time they actually give you these values however I wanted to include an example where they don't give you them just so you can see where these numbers come from in which case just in case your mock or your real thing just happens to not give you the values I don't want you being caught out with your pants down so to speak now FX2 might look a bit tricky but all it is is f * x^2 so quite literally it's just going to be 1 * 2.5 2 and that's it 6.25 and you do this for all of the other values just like you did in the first one so here the frequency is N9 and again we can just change these numbers make it 7.5 gives you 50625 I will show you a slightly faster way of doing this but at least once do it the kind of longer tedious way just so then you can actually properly appreciate the slightly faster way of doing things and this one is 12.5 gives us 29 okay that's quite a big number 29 68.75 296 8.75 and then we have 14 multiply by 20 squ that's going to be pretty huge number so you are you should expect the FX squ to be quite large that's completely fine nothing wrong with that at all and last we have 7 * 32.5 squared so we have seven and we have 32.5 all squared gives me 73 93.75 7 3 93.75 add them all up so once again start from the top so 6.25 + 50625 + 29 68.75 + 5600 + 73 93.75 another good technique by the way is to count how many numbers you have here so we have five and then count how many you have in your calculator missing one is very very common so 1 2 3 four five add those all up and we get a pretty hefty number 22 493 blah blah blah so I'm actually going to put it underneath the box cuz my box is not big enough 4 93.75 and now it's just a matter of subbing it into this formula okay so I'm actually going to just do it all on the calculator so press square root fraction button you have sum of fx^ squ which is going to be this bottom number here cuz remember we've added them all up so 22 493 75 over n² so it's over 50 s then you are subtracting Open Bracket the sum of f ofx Over N once again you should hopefully realize that this is just the mean so if you've worked at the mean in the previous question you can just sub that in squared but just for this case 815 over 50 and that answer is all squared and and that gives us a 1 second something went wrong let me just check it out okay so I spotted the slight mistake uh I actually put an n s here and it's meant to just be n so the actual formula is just n so that was my mistake um I think I was just trying to think ahead but either way the actual formula is the sum of fx^ 2/ N minus the sum of f ofx n all 2 and that does give you the correct value of 13.6 okay so that was a two mark question not too bad however we have a three mark question here which basically says calculate the mean and the sorry the median and the standard deviation of the original data and that is actually quite interesting because here just remember all of these are coded so they have a special they've had this done to them they've subtracted 20 from every value and then halfed it so now we got on to how the mean the median and the standard deviation are affected by this coding because to uncode this let's look at this so we have X is equal to tus 20 over 2 and what you might say is okay let's rearrange for T and that would be a smart move so times both sides by 2 gives me 2x = t - 20 and therefore T is = to 2x + 20 and then you might say okay all I need to do is sub in both of these values into this formula and and we get the original data so remember this is the median of X and the standard deviation of X but we want the median and standard deviation of T the uncoded data and that is half right so for the median and here are some notes for you it's affected by times divide add and subtract and hopefully this makes sense so let's say if I had a bunch of data and it was like 1 3 5 and then actually make the numbers even more simpler so we so we don't have any confusion whatsoever so have 1 2 3 four five okay now you might say the median is three but if I double all of these numbers right so we have two 4 6 8 10 the median also doubles right so if I times them all by two the same would happen if I divide it right the median would go back to three so it's affected by times and divide makes sense right but what if instead I let's say added 10 to every value yeah then once again the median changes and how does the median actually change well the median follows the same pattern if I double all of the values the median doubles and and if I add 10 to all the values the median gets added by 10 and if I divide or subtract the same thing happens because times and uh subtract are just the times and divide are just opposit add and subtract are just opposite so they affect each other the same way so in this case with my median since I've subtracted 20 and divided by two the actual median is going to be 20 more and double so the median for this data is indeed two time 13.9 + 20 just using this exact formula to go from the UN the coded median to the uncoded median and let's just quickly do this on the calculator so 13.9 plus 20 and that gives us 47.8 that makes sense to us I hope so the actual median is 47.8 minutes okay what about the standard deviation well I told you that the standard deviation is a measure of spread so what's another way of measuring spread is the range right well if I have a look at these values here how does the range change well in this first example the spread of the data if used the range is 5- 1 which is four and what about in this data well it's 15 - 11 which is still four the data is still as spread out as it normally is so adding and subtracting does not affect the standard deviation the standard deviation just tells you how spread out the data is it doesn't matter if you shift the data left or right it will not be affected so SD standard deviation is only affected by times divide only but again if we look at this example if we times them by two what happens to the range well 10 - 2 is 8 well the range is twice as big again with this particular example range and standard deviation is effectively the same they tell you about spread with standard deviation it just tells you how far on average the values are from the median okay but we can use the range not for actual specific numbers but as a kind of an analogy okay so because the range is unaffected by adding and subtracting it means all measures of spread are unaffected by adding and subtracting whether that's inter quarar range or standard deviation so when we do the standard deviation when we work at the real standard deviation we ignore the plus 20 instead the standard deviation is just going to be twice as big okay which is going to be 27.2 me I hope that makes sense but keep in mind you don't technically need to understand this you just need to remember it but as I've said in all the videos up till now whether they're the pure statistics mechanics whatever it is so much easier for a level maths if you understand something rather than just memorizing it because memory is not perfect you will make a mistake at some point so go through that one more time if it still doesn't make sense feel free to leave a comment feel free to email leave a message whatever and if it comes to it I'll schedule a video call with you and I'll go through it again but effectively T adding and subtracting doesn't change how spread out the numbers are it only affects how the only thing that affects how spread out the numbers are is timesing and dividing so keep that in mind and that brings us to the end of the first ever standard deviation en coding question so what I'd like you to do is have an attempt at the next question and then watch the video again to uh see what the answer is but I'll see you in the next question okay second part hopefully you guys actually had an attempt now if you notice the first thing I'm spotting and I'm hoping you spotted it too is these numbers do not overlap which means we need to change what the actual class wits are so that way there is actual overlap so please never forget this like I said I have A2 students year 13s that are doing some complicated maths that forget this very basic rule which is another reason why you need to keep revisiting previous topics it may be easy but I guarantee you there's one small little thing you forget about and you know it screws you over now thank God have a look at this they've given us sum of F ofx and some of FX s which means we do not need to do it thank God so they are kind of nice in that case however use linear interpolation to estimate the median height again this is from a previous uh from the question that we did last time but I hope you had a go at doing this don't skip it use every opportunity you can to revise previous topics and if you get the answer wrong there you go there's an early warning system for you go back over it so linear interpolation how do we do it first thing we do is take the number of values and we half it what's the number of values 100 people or add up the frequencies if you're nice and paranoid like I am so that's 50 values the 50th value so if we add this as we're going up we have 4 23 and then I can see straight away that it's going to be in this class over here so what do we do we take the difference in the class boundaries so 29.5 or we just take the class width if you want to say it in a bit more of a human way over the 50th value or in this case we're looking at Q2 minus the lower class width and that equals the difference in the cumulative frequencies or just equal to 41 whichever one's your preference over the actual value we want which is the 50th value minus 23 and once again just do some movie Magic with your value is so if we just first work out what these are so the one on the left is actually quite easy I don't need to do that Q2 - 19.5 work out what the thing is on the right just to make our lives slightly easier 64 - 23 over 50 - 23 41 over 27 nice and easy again you can call it cross multiplying if you like whatever you prefer or you can do it in kind of two different steps so first times up the denominator of the left hand side which gives you Q2 - 19.5 and then we're going to Times by 27 which gives us 270 and divide by 41 = Q2 - 19.5 which means the final answer is going to be Q2 = 270 over 41 + 19.5 nice and lightning fast two marks again it shouldn't take you that long so they do have quite high standards in terms of how quickly you write all this stuff out 26.1 I'm going to call it so when in doubt it's nice to just do three significant figures and then Parts B and C this is so quick this is free easy marks okay estimate the mean height remember what the mean is normally what you do is you do frequency times the midpoint add them all up but that is the same as this so this is another reason why notation very important some students would waste time working out what the midpoint times this is adding it all up and then realizing it's just 2651 well done you've wasted a minute maybe two all it is is the sum of f ofx so the mean is the sum of f ofx over the sum of f is the technical way of writing it but sometimes that confuses people so 2651 over the 100 people here so I'm not going to use a calculator 26.5 to 36 fig which means lastly but not least standard deviation is a Formula you are given but you know remembering it doesn't hurt either sum of fx^ 2 over nus the sum of FX Over N or squ and like I said they've given us these values so so this is nice and easy you could also use this mean in here squared but I do prefer entering the whole thing again even though it doesn't seem like the smart thing to do so 804 so 804 34.25 over this n which was 100 minus 2651 2651 over 100 squared 10.1 let's call it so that was six marks in total and if you have a look at how long it's taken us it's not six minutes and you could probably do it even faster once you start getting really good at it the mean and standard deviation should not take you 4 minutes to be honest at the moment it's okay if it's taking you more than 4 minutes just putting that out there but by the time of the exam this is going to be quick fire nice easy marks that you won't make mistakes on hopefully and now we come on to doing the mean and the standard deviation for ungrouped data so this is again the the main reason why I wanted to put this in here is just to illustrate how the formulas change so we know that the mean is the sum of f ofx over n and then we have the standard deviation formula as well but what happens when we don't have a midpoint X well if you actually think about what this is here the midpoint it's the actual value of the data right the fre frequency isn't the value of the data the actual value of the data here is the midpoint so the act the thing that disappears isn't X the midpoint it's F that disappears okay because we don't have a frequency it's not that we don't have the data we don't have the frequency so with the mean here all we're going to do is we're going to add up all of the pieces of data and divide it by the number of data we have and I like to do this all in kind of one step on a calculator so top is just going to be adding all of these up so 45.2 46.9 46.1 46.2 45.4 45.1 47.8 uh lost my space 45.4 45.5 46.1 46.4 and 45.6 seven so let's just go through and make sure we get it do it backwards so one key thing is not to read through the numbers in the same order again so one thing they actually use they you tell snipers to do this is to go in the opposite way to what you normally read so we normally read left to right top down if you want actually spot mistakes you read right to left and down up so we can't really do down up here but we read it right to left so 45.7 46.4 46.1 45.5 45.4 47.8 45 .1 45.4 46.2 46.1 46.9 and 45.2 and what it does it forces your brain to actually pay more attention and we have 12 athletes let's just divide by 12 there we go so 45.9 so we should probably round this to three significant figures so 46.0 to 3 Sig fig now what about for the standard deviation once again it's just going to be the exact same formula but we just delete the fs so how is that going to look let me show you so it's a sum of x^2 Over N minus the sum of x/ n^ 2 okay so here we're probably going to have to do this separately but what we can do if we're quite clever is we can actually use this so what we're going to do is we want to delete the fraction and that's just the sum of X so now we have the sum of X is equal to 55 1.8 okay and how do we get the sum of X2 well all I need to do is go through and put a squared sign after each digit so there squared there squared squ actually I don't know why I'm using and if you actually remember the spacing between each number because they're all the same number of digits you can actually do this a lot faster so you just go I think it's what would it be uh it' be one two three four so five to the right so one two three four five boom one two 3 4 5 1 2 3 4 oh never mind I got to the end uh 25 38. 38 so 25 38. 38 and then we just need to sub it in nice and easy hopefully possibly so we have the sum of X squ so 25 38038 over the 12 athletes minus 55 1.8 over 12 and that is squared again 0.75 7.5 0.51 like that 0.751 to 3sf so that's not too bad hopefully so far it's uh been okay but again I would like you to have a go at the next question before you kind of uh look at the video if you can okay and for the final questions the final kind of type of question is about the l l the large data set that they may give you and the data can look a bit confusing because it's real world data which is tends to be a bit messy but it's the exact same stuff that we've been doing for the past few videos including this one and as you may notice there is no part A and there's a few other parts missing because once again large data set question has normally four or five parts but they outside the scope of this video first thing we need to do in this particular case is use linear interpolation to calculate an estimate for the upper quartile so yeah linear interpolation is not just for median it's for the upper and lower quarts as well you just need to remember the different formulas for all of them so upper quartile is 3 n over 4 and the lower quartile is just n over4 so let's do it we need to work out what the frequency is however what we can do is add all them up right we don't it's not the hardest thing in the world 12 + 17 so it's 184 so now if we put this into our calculation 3 * 184 over 4 and what I'm going to do because I know that I'm going to forget is I'm going to write that the sum is 184 because that is going to be useful a bit later on we're looking for the 138th value it's actually not as hard as it seems cuz this is already at 121 it's 131 that's going to be 155 yeah so it's definitely in this range here so that's nice and easy so once again we're going to do the difference in the class width 5 -1 over Q3 -1 is equal to difference in the cumulative frequencies 155 - 131 over 138 - 131 again nice and easy nice and standard so we have 4 over Q3 minus 1 equals and then whatever the heck that is on the other side so we have 155 minus 131 over 138 - 131 24 7 okay it's a bit simplified that's nice actually no it isn't what am I talking about do the good old fashioned times into the other side we get 4 is equal to 24 7 Q3 -1 I would like to point out that since your algebra is going to get a lot better over these next 2 years or until you finish A2 anyway you will be able to do this a lot faster than I'm showing you right now because I'm taking it nice and slow just on the assumption that you might still be slightly weak which is okay it's okay to be weak so it's not okay to stay that way 7 * 4 over 21 add another one for good luck and that will be actually just put the equals on the side just to keep it nice and neat we have 7 multiplied 4 divided 21 add another one 7 over three so that's 2.33 to three significant figures and with the standard deviation we already know it nice and easy all we need to do is we're going to use Sigma equals write out for the last time in this video sum of f of x^2 Over N minus the sum of FX Over N all squar like that and they give us the values look how lovely and kind the examiners are so 7704 7704 1875 Over N which I said was 184 perfect minus 539 75 39.75 over 184 squared 5.77 and the number being small should hopefully make some amount of sense because it is rainfall okay so it does make sense for it to be a relatively small spread and that's it for standard deviation and coding it doesn't really get much harder than that with the coding I would recommend finding problems to try and practice so going back to here with the coding times but it really is just this if you know this if you're comfortable with this you can do coding it really does not ever get harder than that it's nice and easy it affects the median in all ways times divide add and subtract and it only affects standard deviation by timesing and dividing it's pretty simple so again if you find another example should be nice and easy but again the mark scheme is going to be online the work she's going to be online